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A329333
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There is exactly one odd prime among the pairwise sums of any three consecutive terms: Lexicographically earliest sequence of distinct nonnegative integers with this property.
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40
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0, 1, 2, 7, 3, 6, 4, 5, 8, 10, 11, 9, 12, 14, 15, 13, 18, 17, 19, 20, 21, 24, 16, 23, 25, 22, 26, 27, 28, 31, 29, 32, 33, 34, 30, 39, 37, 36, 38, 41, 40, 42, 43, 46, 35, 44, 47, 45, 50, 51, 48, 49, 56, 52, 53, 54, 57, 55, 58, 59, 68, 60, 63, 64, 61, 66, 62, 69, 67, 72, 71, 65, 74, 70, 75, 76, 77
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OFFSET
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0,3
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COMMENTS
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This is conjectured and designed to be a permutation of the nonnegative integers, therefore the offset is taken to be zero.
Restricted to positive indices, this is a sequence of positive integers having the same property, then conjectured to be a permutation of the positive integers. (The word "odd" can be omitted in this case.)
If the word "odd" is dropped from the original definition, the sequence starts (0, 1, 3, 6, 2, 7), and then continues from a(6) = 4 onward as the present sequence. This is again conjectured to be a permutation of the nonnegative integers, and a permutation of the positive integers when restricted to the domain [1..oo). The latter however no longer has the property of lexicographic minimality.
See the OEIS wiki page for further considerations about existence, surjectivity and variants. - M. F. Hasler, Nov 24 2019
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LINKS
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EXAMPLE
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For the first two terms there is no restriction regarding primality, so a(0) = 0, a(1) = 1. (If only positive values and indices are considered, then a(1) = 1 and a(2) = 2.)
Then a(2) must be such that among { 0+1, 0+a(2), 1+a(2) } there is exactly one odd prime, and 2 works.
Then a(3) must be such that among { 1+2, 1+a(3), 2+a(3) } there is only one (odd) prime. Since 1+2 = 3, the other two sums must both yield a composite. This excludes 3, 4, 5 and 6 and the smallest possibility is a(3) = 7.
And so on.
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MATHEMATICA
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a[0]=0; a[1]=1; a[2]=2; a[n_]:=a[n]=(k=1; While[Length@Select[Plus@@@Subsets[{a[n-1], a[n-2], ++k}, {2}], PrimeQ]!=1||MemberQ[Array[a, n-1, 0], k]]; k); Array[a, 100, 0] (* Giorgos Kalogeropoulos, May 07 2021 *)
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PROG
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(PARI) A329333(n, show=0, o=0, p=0, U=[])={for(n=o, n-1, show&&print1(o", "); U=setunion(U, [o]); while(#U>1&&U[1]==U[2]-1, U=U[^1]); for(k=U[1]+1, oo, setsearch(U, k)|| if(isprime(o+p), isprime(o+k)|| isprime(p+k), isprime(o+k)==isprime(p+k)&&p)||[o&&p=o, o=k, break])); o} \\ Optional args: show = 1: print all values up to a(n); o = 1: start with a(1) = 1; p = 1: compute the variant with a(2) = 3. See the wiki page for more general code which returns the whole vector: Use S(n_max, 1, 3, 1) or S(n_max, 1, 3, 2, [0, 1]); S(n_max, 1, 3, 0) gives the variant (0, 1, 3, ...)
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CROSSREFS
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See A329450 for the variant having 0 primes among a(n+i) + a(n+j), 0 <= i < j < 3.
See A329452 for the variant having 2 primes among a(n+i) + a(n+j), 0 <= i < j < 4.
A084937, A305369 have comparable conditions on three consecutive terms.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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